Quantum Gravity Seminar

Week 1, Track 1

Toby Bartels

October 2, 2000

From deep within the Wizard's lair came a cackle of glee.

"At last!" cried the Wizard "I have assembled a Quantum Gravity Crew.
Together, we will take over the world -- or at least learn more about it.
Now I can gather them together for the next step in my plan: --"
Here an ominous crash of thunder broke into his reverie.
He continued "-- A SEMINAR!!! BWAHAHAHAHA!!!!!!!!!"
A tad embarrassed by his excessive use of punctuation,
the Wizard emerged from his cave and made his way to the seminar room.

There the Quantum Gravity Crew waited with unconcealed delight.
But there was a problem. Several naive mathematics students
had also signed up for the seminar in hopes of an easy A.
Without knowledge of GR and QFT, would they be able to hack it?

But they don't call the Wizard "the Wizard" for nothing.
"This seminar will consist of 2 tracks," the Wizard announced
"like Misner, Thorne, and Wheeler, or, for that matter, the phone book.
Track 1 will require only linear algebra,
including tensor products and dual spaces.
Track 2 will require familiarity with differential geometry,
and it's good to know some general relativity and quantum theory.
Let's begin with ... track 1.". Relieved, the class sighed.

"The great quest behind quantum gravity is the reconciliation
of 2 great theories of the universe that are completely different:
general relativity (GR) and quantum mechanics (QM).".
Here the Wizard waved his wand, and a diagram appeared on the board:

___ ____
/ \___/ \
| __ \ space
| / \___ |
\_/ \___/

"General relativity describes space as a manifold.
Don't worry if you don't know what a manifold is;
it's just a smooth geometric thing with a dimension.
Here, the manifold is a 3 dimensional space which describes
how the world could be at one instant of time.

Now, at another instant in time, space could be something else.".
And the Wizard waved his wand again, a bit lower down this time.

___ __
/ \_____/ \
| \ space
| ___ |
\_____/ \___/

"But we know that space is not the whole story.
There is also spacetime, which a 4 dimensional manifold,
which we can draw like this:

Here the Wizard paused to catch his breath.
Jay, the Acolyte of Physics, interrupted with a question.
"But isn't it arbitrary how you divide spacetime
into different instants of time to have space in?".

"True -- but not important just yet." replied the Wizard
"I'm only motivating an idea now.".

"On the other hand," continued the Wizard "we have quantum mechanics.
In this case, we have a complex Hilbert space,
which is a kind of vector space with an inner product.
It doesn't draw as well, but I can give it a symbol, "H".
It's now a normalized vector in this Hilbert space
which represents a state, a way the world can be at a given time.
Now we have, to represent the passage of time,
a time evolution operator U, which goes between Hilbert spaces.
It looks like this:

H Hilbert space
|
|
|
U operator
|
|
|
v
H' Hilbert space

The mathematical language for this is functional analysis,
which is linear algebra with a bit of analysis thrown in.

Differential geometry and functional analysis are totally different things!
But the 2 diagrams on the board have something in common:
they both describe states, the way the world could be,
and processes going between states, the way the world can change.

Toby, the Acolyte of Mathematics, smiled inwardly to himself,
as visions of ncategories danced in his head.

"Let's start with diagrams of operators on vector spaces." said the Wizard
"By "vector space", I mean a finite dimensional complex vector space.
Suppose we have a linear operator T from the space V to the space W.
Now, we could write this as T: V -> W, but I'm going to do it differently.".
And, waving his wand:

|
|
V v
|
|
/ \
| T |
\_/
|
|
W v
|
|

"You can think of this as a bit of pipe with a machine in it.
An element v of V falls through the pipe labeled "V".
Then it enters the machine labeled "T" and comes out
as an element of W falling through the pipe labeled "W".

Now, one thing we can do with linear operators is to compose them.
If T: V -> W and S: U -> V, then TS: U -> W is the composition.
We can draw TS in either of 2 ways. We can separate it as:

|
|
U v
|
|
/ \
| S |
\_/
|
|
V v
|
|
/ \
| T |
\_/
|
|
W v
|
|

or put it all together into one machine as

|
|
U v
|
|
/ \
| TS|
\_/
|
|
W v
|
|

These pictures represent the same thing.
By saying that they represent the same thing,
I am defining what the first of the 2 pictures means.

Does this mean (TS)R or T(SR)? It's ambiguous!
Except it's not really ambiguous, because composition is associative,
and (TS)R *equals* T(SR).

So, you see what we're really trying to do with these diagrams.
We're trying to make things which are *theorems* in real life
seem perfectly obvious in diagrammatic language.

Now, I have a question. If we have any vector space V,
what linear operator do we get automatically, for free?"

Miguel, the Acolyte who Can't Decide if he's Physics or Math, said
"How about the identity function?".

"Right!" said the Wizard "The identity function 1V is a linear operator.
We draw it like this:

|
|
V v
|
|

That's it. Just a piece of pipe that elements of V can fall through.
Nothing happens to the elements when they fall through,
so it's just as if there's no machine, so we don't draw one.

Now, there's another possible ambiguity. We know that

|
|
V v
|
|
/ \
| T |
\_/
|
|
W v
|
|

is the picture for the operator T.
But how do we know if the pipe above has the identity for V in it?
And how do we know if the pipe below has the identity for W in it?
In other words, is this picture T, T(1V), or (1W)T?
The solution, of course, is that these are all the same.
So, again, a possible ambiguity in the diagram
becomes an obvious expression of a theorem.".

"Now," said the Wizard, deftly slipping through time,
"let me consider dual spaces and the dual of a linear map.
If V is a vector space, then the space of all linear maps
from V to the complex numbers C is itself a vector space.
This vector space is V*, the dual of V.
How do we draw this space (or the identity map on this space)?
Well, we could draw it like this:

|
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V* v
|
|

But I'm going to draw it like this instead:

|
|
V ^
|
|

See, when the arrow on the pipe goes upwards, it means the dual space.
So, you have to keep track of up and down in your drawing,
so you know which way the arrow is pointing.

Now, if you have a linear map T from V to W,
then you get a dual map T* from W* to V*.
Specifically, if L: W -> C and v in V,
then let (T*L)v be LTv. Then T*L is in V*.

How do we draw this? Well, we know W* needs to be on top,
even though W was on bottom and V on top before.
Also, we know the arrows need to point upwards instead of downwards.
You can see what we have to do: turn everything upside down.
And that's our general rule for duals:
To get the dual of something, turn it all upside down.
The diagram for T*, then, is:".

|
|
W ^
|
|
/ \
| T*|
\_/
|
|
V ^
|
|

Oz, who I'll stick into this story on the grounds that
he is a recurring character in the Wizard's exploits,
tentatively raised his hand. The Wizard prepared a fireball.

"Do we really need to indicate the "*" in the diagram?
Won't the upward arrows tell us that it's T* instead of T,
just like they tell us it's V* instead of V?".

Foom! Oz took on the appearance that Elmer Fudd
usually takes on just after he attempts to shoot Bugs Bunny.

"Do you think I'd put it in my notation if it wasn't necessary!?"
shouted the Wizard, trying to hide the fact that, technically,
it wasn't necessary to label the pipes "V" and "W"
since the operator T knows what its domain and codomain are,
"What if T is originally an operator from V to V*?
Then T* is also an operator from V to V*, since V** = V,
but T* isn't necessarily the same operator as T.
Then the notation

|
|
V v
|
|
/ \
| T |
\_/
|
|
V ^
|
|

would really be ambiguous! We don't want that.".

(Actually, this conversation never happened,
but I need to settle this point for my own sake,
and it's always cool to pick on Oz.
Otherwise, the fireball would get thrown at me,
and that's no good.)

"OK," said the Wizard after he calmed down
"now let's consider tensor products.
Since our vector spaces are finite dimensional,
we can define V (x) W as the space of all linear maps from W* to V.
(In infinite dimensions, we'd have to be more sneaky.)
Again, we can form tensor products of linear operators.
If S: U -> V and T: W -> X, then S (x) T: U (x) W -> V (x) X;
to define it, if R: W* -> U is an element of U (x) W
and L: X -> \C is an element of X*,
then let ((S (x) T)R)L be SRT*L, which is in V.
Then (S (x) T)R is an element of V (x) X,
and S (x) T is a linear operator from U (x) W to V (x) X.

That's all a mouthful, but it's easy to draw.
We just put S and T side by side:

I don't remember who said this, so let's make it Oz again.
"Well, it seems to me that the S comes before the T.
So, really, the S is beside the identity on W,
and the T is beside the identity on V.
So, I guess that would make it (S (x) 1W)(1V (x) T) --
No, wait! You had us write composition the other way,
so it's (1V (x) T)(S (x) 1W).".

Trying his best to hide his disappointment,
the Wizard put away the fireball he had been preparing.
"That's absolutely correct!" he said
"Now, the problem is that we don't want to always have to
look at our picture carefully to find out
which of a pair of things comes before the other
or whether they're happening at the same time.
Luckily, we don't have to! It's another obvious theorem:
(1V (x) T)(S (x) 1W) = S (x) T.
More generally, (Q (x) R)(S (x) T) = QS (x) RT.
That's obvious when you look at the diagram

Here, T is an operator from U (x) W to V* (x) X,
S is an operator from X (x) Z* to Y,
and the picture shows a certain way to compose them
to get an operator from U (x) W (x) Z* to V* (x) Y.".

"OK!" said the Wizard "I've already gone over time for track 1.
Now we go on to track 2, with Lie algebras and differential forms.".

Screaming, Oz ran from the room. Half the class followed him.

"Now that Oz has left," said Toby the Acolyte
"can I make a short category theoretic note?
It seems to me that, when you draw composition,
you only need a 1 dimensional diagram -- nothing to the side.
And this sort of diagram can be drawn for any category,
not just the category of finite dimensional complex vector spaces.
Then, when you introduce tensor products, you get 2D diagrams,
and this can be drawn for any monoidal category.
If the monoidal category has a braiding,
you can cross lines in the diagram, one behind the other,
which takes 3 dimensions, so braided monoidal categories
have 3D diagrams. Finally, if the braiding is symmetric,
then it doesn't matter which way you cross lines,
which indicates that you really have a 4th dimension
to switch things around in, so symmetric monoidal categories
have 4D diagrams. And turning things upside down
works in any [[[symmetric | braided]] monoidal] category with duals.".

"Yes." said the Wizard.

"Toby," said Miguel the Acolyte "can you write that down,
so I can put it in the notes I'm making for
Nathan, the Acolyte who Went to the Wrong School, Poor Sucker?".

"No problem." said Toby the Acolyte
"I'll put it in the summary I write for s.p.r.".